Online Test For Reflexive Binary Relation

Online test for reflexive binary relation

You are correct about reflexivity and symmetry.

Online test for reflexive binary relation

The main diagonal is all $1$’s, so the relation is reflexive, and the matrix is symmetric about the main diagonal, so the relation is symmetric.

A relation $R$ is antisymmetric if $$(a R b\text{ AND }bRa)\implies (a=b)\;.$$ For your first relation you have $1R3$ and $3R1$, and yet $3\ne 1$, so this relation is not antisymmetric.

For the same reason the second and fourth relations are not antisymmetric.

Can an llc do an ipo

The third relation is not antisymmetric because $1R5$ and $5R1$, even though $1 \ne 5$. The fifth is not antisymmetric because $2R5$ and $5R2$, even though $2 \ne 5$.

Online test for reflexive binary relation

Thus, none of the relations is antisymmetric. The visual test is to see whether there is a pair of $1$’s that are symmetric with respect to the main diagonal; if there are $-$ and we found such a pair of every one of these five relations $-$ then the relation is not antisymmetric.

There is no simple test for transitivity.

For the first relation, note that $2R3$ and $3R1$, but it’s not true that $2R1$, so the first relation is not transitive.

Online test for reflexive binary relation

In the third relation $1R4$ and $4R2$, but it’s not true that $1R2$, so the relation is not transitive. In the fourth relation $2R3$ and $3R1$, but it isn’t true that $2R1$, so the relation is not transitive.

Online test for reflexive binary relation

In the last relation $3R2$ and $2R1$, but it’s not the case that $3R1$, so the relation is not transitive.

The second relation is, as you say, transitive. One way to demonstrate this is to notice that in this relation every odd number in $\{1,2,3,4,5\}$ is related to every odd number, and every even number to every even number, but no odd number is related to any even number or vice versa.

Binary Relations IIT JEE Main Mathematics Online Mock Test Problem 61(April 15 2018 Morning)

Thus, if $aRb$ and $bRc$, then either $a,b$, and $c$ are all odd, or they’re all even, and in either case it’s true that $aRc$.

$\endgroup$

answered Dec 30 '11 at 11:21

Brian M.

ScottBrian M. Scott

474k4242 gold badges568568 silver badges973973 bronze badges

Fidelity online trading options